1.

The 'activation energy' required for a biochemical reaction in a cell is drastically reduced by
enzymes. This is primarily achieved by: (A) Forming a permanent enzyme-substrate complex
(B) Increasing the temperature of the system (C) Providing a different path with a lower energy
transition state (D) Altering the free energy change (”G) of the reaction

2. A student is given four samples of monosaccharides: I. Glyceraldehyde, II. Ribose, III.

Fructose, IV. Galactose. Which of these are ketoses and aldoses respectively? (A) I & II are
aldoses; III & IV are ketoses (B) I, II & IV are aldoses; III is a ketose (C) I & IV are aldoses; II &
III are ketoses (D) II & III are aldoses; I & IV are ketoses

3. The primary structure of a protein is directly determined by: (A) Hydrogen bonding between
amino acids (B) The sequence of codons on mRNA (C) The sequence of anticodons on tRNA
(D) Ionic interactions within the polypeptide chain

4. In a DNA molecule, the ratio of which of the following is 1 : 1, according to Chargaff's rule?
(A) A : T and G : C (B) A : G and T : C (C) A : C and G : T (D) (A + T) : (G + C)

5. The cofactor 'Haem' in haemoglobin is: (A) A prosthetic group (B) A coenzyme (C) An
inorganic ion (D) An apoenzyme

6. Which of the following statements is INCORRECT regarding polysaccharides? (A) Inulin is

a polymer of fructose and is used to determine renal function. (B) Cellulose is a homopolymer
of glucose with ²(1!’4) glycosidic linkages. (C) Glycogen has a highly branched structure with
both ±(1!’4) and ±(1!’6) linkages. (D) Chitin is a heteropolysaccharide found in the exoskeleton
of arthropods.

7. A nucleic acid sample was analyzed and found to have 22% Adenine, 28% Guanine, 28%
Cytosine, and 22% Uracil. The molecule is most likely: (A) Double-stranded DNA (B)
Double-stranded RNA (C) Single-stranded DNA (D) Single-stranded RNA

8. The 'lock and key' model of enzyme action differs from the 'induced fit' model in that it: (A)
Proposes a rigid, pre-shaped active site (B) Explains the formation of the enzyme-substrate
complex (C) Involves a change in the shape of the enzyme (D) Accounts for the broad
specificity of some enzymes

9. Which of the following is NOT a secondary metabolite? (A) Ricin (B) Curcumin (C) Leucine
(D) Vinblastin

10. Competitive inhibition of an enzyme can be overcome by: (A) Removing the inhibitor by
dialysis (B) Adding more substrate (C) Adding a non-competitive inhibitor (D) Increasing the
pH

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Answer Key & Detailed Explanations

1. Answer: (C) Providing a different path with a lower energy transition state
· Explanation: Enzymes work by stabilizing the high-energy transition state of the reaction,
effectively creating a new reaction pathway that requires less activation energy. They do not
change the ”G of the reaction (D is wrong) and the ES complex is not permanent (A is wrong).
2. Answer: (B) I, II & IV are aldoses; III is a ketose
· Explanation: Glyceraldehyde (3C), Ribose (5C), and Galactose (6C) are aldoses (have an
aldehyde group). Fructose (6C) is a ketose (has a ketone group).
3. Answer: (B) The sequence of codons on mRNA
· Explanation: The primary structure is the linear sequence of amino acids. This sequence is
coded for by the sequence of codons on mRNA. While tRNA brings the amino acids, the code
itself is on mRNA.
4. Answer: (A) A : T and G : C
· Explanation: Chargaff's rules state that in double-stranded DNA, the amount of Adenine
equals Thymine (A=T) and the amount of Guanine equals Cytosine (G=C). Therefore, their
ratios are 1:1.
5. Answer: (A) A prosthetic group
· Explanation: Haem is an organic, non-protein cofactor that is tightly and permanently bound
to the apoenzyme (globin in haemoglobin). It is not loosely bound like a coenzyme (B) and is
organic, not inorganic (C).
6. Answer: (D) Chitin is a heteropolysaccharide found in the exoskeleton of arthropods.
· Explanation: This is the incorrect statement. Chitin is a homopolysaccharide, a linear polymer
of N-acetylglucosamine. The other statements are correct.
7. Answer: (D) Single-stranded RNA
· Explanation: The presence of Uracil (U) instead of Thymine (T) confirms it is RNA, not DNA.
Chargaff's rules (A=T, G=C) do not hold here (A=22%, but T is 0%; U=22%). The equal
percentages of A and U, and G and C, are a coincidence often seen in single-stranded nucleic
acids that form hairpin loops, but the key is the presence of U and the violation of base-pair
equality rules, indicating it is single-stranded.
8. Answer: (A) Proposes a rigid, pre-shaped active site
· Explanation: The 'lock and key' model (proposed by Emil Fischer) suggests the active site is
a rigid, pre-formed structure complementary to the substrate. The 'induced fit' model (by
Daniel Koshland) proposes the active site is flexible and molds itself around the substrate.
9. Answer: (C) Leucine
· Explanation: Leucine is an essential amino acid, which is a primary metabolite involved in
fundamental physiological processes. Ricin (toxin), Curcumin (pigment), and Vinblastin
(alkaloid, anti-cancer drug) are all secondary metabolites.
10. Answer: (B) Adding more substrate
· Explanation: In competitive inhibition, the inhibitor competes with the substrate for the active

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site. By increasing the substrate concentration, the probability of a substrate molecule binding
to the enzyme instead of the inhibitor increases, thus overcoming the inhibition.

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